$2xy+x^3-3y^2=5$ Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{2x-6y}{2y-3x^2}$ (Choice B) B $-\dfrac{2y+3x^2}{2x-6y}$ (Choice C) C $\dfrac{-3x^2}{6y-2}$ (Choice D) D $\dfrac{6y-3x^2}{2 }$
Explanation: We cannot isolate $y$ in order to define it as a function of $x$. Therefore, $2xy+x^3-3y^2=5$ defines $y$ as a function of $x$ implicitly. To find $\dfrac{dy}{dx}$, we need to perform implicit differentiation. In implicit differentiation, we differentiate both sides of the equation according to $x$, and treat $y$ as an implicit function of $x$. [I need more explanation about implicit differentiation!] $\begin{aligned} 2xy+x^3-3y^2&=5 \\\\ \dfrac{d}{dx}(2xy+x^3-3y^2)&=\dfrac{d}{dx}(5) \\\\ 2\dfrac{d}{dx}(xy)+\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3y^2)&=0 \\\\ 2\Bigl(1\cdot y+x\cdot\dfrac{dy}{dx}\Bigr)+3x^2-6y\cdot\dfrac{dy}{dx}&=0 \\\\ 2y+2x\cdot\dfrac{dy}{dx}+3x^2-6y\cdot\dfrac{dy}{dx}&=0 \end{aligned}$ Once we've completed the differentiation, we can arrange the equation so $\dfrac{dy}{dx}$ is isolated: $\begin{aligned} 2y+2x\cdot\dfrac{dy}{dx}+3x^2-6y\cdot\dfrac{dy}{dx}&=0 \\\\ \dfrac{dy}{dx}(2x-6y)&=-(2y+3x^2) \\\\ \dfrac{dy}{dx}&=-\dfrac{2y+3x^2}{2x-6y} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{2y+3x^2}{2x-6y}$.